if cosec A - sin A=a 3
, sec A - cos A=b 3
,prove that a2
b
2
(a2
+b2
) = 1
priyadharsan:
i can't understand the 'prove that' part
can u comment it?
a2b2(a2+b2)=1
hope it helps
Answers
Answered by
22
cosec A - sin A = a^3
= (1/sin A) - sin A
= (1 - (sin A)^2)/sin A
= ((cos A)^2)/sin A
= cot A . cos A = a^3
similarly,
sec A - cos A = tan A . sin A
= tan A . sin A = b^3
(a^2 b^2) (a^2 + b^2) = (((ab)^3)^(2/3)) ((a^3)^(2/3) + (b^3)^(2/3))
[ (cot A cos A tan A sin A)^(2/3) ] [ (cot A . cos A)^2/3 + ( tan A . sin A)^2/3) ]
tan A and cot A get cancelled.
Remaining steps are in the attachment.
= (1/sin A) - sin A
= (1 - (sin A)^2)/sin A
= ((cos A)^2)/sin A
= cot A . cos A = a^3
similarly,
sec A - cos A = tan A . sin A
= tan A . sin A = b^3
(a^2 b^2) (a^2 + b^2) = (((ab)^3)^(2/3)) ((a^3)^(2/3) + (b^3)^(2/3))
[ (cot A cos A tan A sin A)^(2/3) ] [ (cot A . cos A)^2/3 + ( tan A . sin A)^2/3) ]
tan A and cot A get cancelled.
Remaining steps are in the attachment.
Attachments:
Similar questions